$\begin{cases}b(1)=-500\\\\ b(n)=b(n-1)\cdot \dfrac{4}{5} \end{cases}$ What is the $3^{\text{rd}}$ term in the sequence?
This is a recursive formula. It tells us that the first term is $-500$ and that the common ratio is $\dfrac{4}{5}$. $\begin{aligned} {b(1)}&=-500 \\\\ {b(2)}&={b(1)}\cdot \dfrac{4}{5}=-400 \\\\ {b(3)}&={b(2)}\cdot \dfrac{4}{5}=-320 \end{aligned}$ The $3^{\text{rd}}$ term is $-320$.